Pure benzene, C₆H₆, freezes at 5.5^$\circ$C and boils at 80.1^$\circ$C. What is the boiling point of a solution consisting of cyclohexane (C₆H₁₂) dissolved in benzene if the freezing point of this solution is 0.0^$\circ$C (For benzene, K_f = 5.12 ^$\circ$C/m, K_b = 2.53 ^$\circ$C/m; for cyclohexane, K_f = 20.0 ^$\circ$C/m, K_b = 2.79^$\circ$C/m) A) 77.4^$\circ$C B) 80.9^$\circ$C C) 82.8^$\circ$C D) 83.1^$\circ$C E) 91.2^$\circ$C

Question

Quizplus · Accepted Answer

First, we need to use the formula for the change in freezing point:

$\Delta T_f = K_f \cdot m$

Where $\Delta T_f$ is the change in freezing point (in this case, -5.5°C), $K_f$ is the freezing point depression constant (for benzene, 5.12°C/m), and $m$ is the molality of the solution.

We can rearrange this formula to solve for the molality:

$m = \frac{\Delta T_f}{K_f}$

Plugging in the values, we get:

$m = \frac{-5.5°C}{5.12°C/m} = -1.074 m$

Note that the negative sign just indicates that the freezing point is lower than the normal freezing point of the solvent.

Next, we need to use the formula for the change in boiling point:

$\Delta T_b = K_b \cdot m$

Where $\Delta T_b$ is the change in boiling point (which we want to find), $K_b$ is the boiling point elevation constant (for benzene, 2.53°C/m), and $m$ is the molality of the solution (which we just calculated).

We can rearrange this formula to solve for the change in boiling point:

$\Delta T_b = K_b \cdot m$

Plugging in the values, we get:

$\Delta T_b = 2.53°C/m \cdot (-1.074 m) = -2.72°C$

(Note that the negative sign just indicates that the boiling point is lower than the normal boiling point of the solvent.)

Finally, we add the change in boiling point to the normal boiling point of the solvent (benzene):

$80.1°C - 2.72°C = 77.38°C$

Rounding to one decimal place, the answer is:

$77.4°C$.

Therefore, the correct answer is C.

Pure Benzene, C6H6, Freezes at 5 ∘\circ∘ C and Boils at 80

Pure Benzene, C₆H₆, Freezes at 5 $\circ$ C and Boils at 80